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## Homework Statement

A box with mass 2kg is on the edge of a circular platform of radius 6.0m. The coefficient of friction between the platform and the box is 0.3. The platform accelerates. Determine the speed when the box slips off the edge.

## Homework Equations

Fs = μsN

F⃗ net=ΣF⃗ =ma⃗

a(t) = d(vt)/dt

a(r) = v^2/r

a = [sqrt (ar^2) + (at^2)]

## The Attempt at a Solution

I know that this is non-uniform circular motion.

I don't think that what I'm doing is right.

ΣF(t) = -Ff = m(at)

at = - μs(g)

a(t) = - (0.3)(9.8)

a(t) = -2.94 m/s^2

ΣF(r) = m(ar)

ΣF(r) = (mv^2) / r

ΣF(net) = m * [sqrt (ar^2) + (at^2)]

Ff = m * [sqrt (ar^2) + (at^2)]

μs(g) = [sqrt (μs^2(g^2)) + (m^2*v^4) / r^2]

(μg)^2 = (μs^2(g^2)) + (m^2v^4) / r^2

((μg)^2 - μs^2(g^2)) /r^2 = (m^2*v^4)

So if I solve for v, I would take the fourth root of the left side after bringing m^2 over, and that doesn't seem right to me.